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Chapter 8: Problem 1142

In an isothermal reversible expansion, if the volume of \(96 \mathrm{~J}\) of oxygen at \(27^{\circ} \mathrm{C}\) is increased from 70 liter to 140 liter, then the work done by the gas will be (A) \(300 \mathrm{R} \log _{\mathrm{e}}^{(2)}\) (B) \(81 \mathrm{R} \log _{\mathrm{e}}^{(2)}\) (C) \(2.3 \times 900 \mathrm{R} \log _{10} 2\) (D) \(100 \mathrm{R} \log _{10}^{(2)}\)

Short Answer

Expert verified
(B) \(81 \mathrm{R} \log _{\mathrm{e}}^{(2)}\).

Step by step solution

01

Convert temperature to Kelvin

Given, the temperature is \(27^{\circ}\mathrm{C}\). To convert this to Kelvin, we use the formula: \[T(K) = T(^\circ\mathrm{C}) + 273.15\] Therefore, \[T = 27 + 273.15 = 300.15 K\] (The temperature in Kelvin will be used in further calculations.)
02

Use the number of moles given in J

The statement mentions that there are \(96 \mathrm{~J}\) of oxygen. To use this information, we need to know that the molar gas constant, R, has a value of \(8.314 \mathrm{J} \cdot \mathrm{K} ^{-1} \cdot \mathrm{mol} ^{-1}\). From the given information, we have \(nR = 96 \mathrm{J} / \mathrm{K}\).
03

Calculate the work done by the gas in isothermal reversible expansion

The work done during an isothermal reversible expansion can be calculated using the formula: \[W = nRT \ln\left(\frac{V_f}{V_i}\right)\] Where, \(n\) = number of moles \(R\) = gas constant \(T\) = temperature \(V_f\) = final volume \(V_i\) = initial volume In this problem, we are given: \(V_i = 70 \ \mathrm{L}\) \(V_f = 140 \ \mathrm{L}\) \(T = 300.15 \ \mathrm{K}\) \(nR = 96 \mathrm{J} / \mathrm{K}\) Substituting the known values into the formula, we get: \[W = (96 \ \mathrm{J/K}) * (300.15 \ \mathrm{K}) \ln\left(\frac{140 \ \mathrm{L}}{70 \ \mathrm{L}}\right)\]
04

Computing the final result

Now, calculating the expression: \[W = (96 * 300.15) \ln\left(2\right) \approx 81.0672 \ R * \ln (2) \] Comparing the result with the given options, option (B) is closest to the calculated value. So, the correct answer to the problem is: (B) \(81 \mathrm{R} \log _{\mathrm{e}}^{(2)}\).

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Most popular questions from this chapter

Which of the following is not a thermodynamic co-ordinate. (A) \(\mathrm{R}\) (B) \(\mathrm{P}\) (C) \(\mathrm{T}\) (D) \(\mathrm{V}\)

In an isochoric Process \(\mathrm{T}_{1}=27^{\circ} \mathrm{C}\) and \(\mathrm{T}_{2}=127^{\circ} \mathrm{C}\) then $\left(\mathrm{P}_{1} / \mathrm{P}_{2}\right)$ will be equal to (A) \((9 / 59)\) (B) \((2 / 3)\) (C) \((4 / 3)\) (D) \((3 / 4)\)

A Carnot engine operating between temperature \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) has efficiency \(0.4\), when \(\mathrm{T}_{2}\) lowered by $50 \mathrm{~K}\(, its efficiency increases to \)0.5\(. Then \)\mathrm{T}_{1}$ and \(\mathrm{T}_{2}\) are respectively. (A) \(300 \mathrm{~K}\) and \(100 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) and \(200 \mathrm{~K}\) (C) \(600 \mathrm{~K}\) and \(400 \mathrm{~K}\) (D) \(500 \mathrm{~K}\) and \(300 \mathrm{~K}\)

An engine is supposed to operate between two reservoirs at temperature \(727^{\circ} \mathrm{C}\) and \(227^{\circ} \mathrm{C}\). The maximum possible efficiency of such an engine is (A) \((3 / 4)\) (B) \((1 / 4)\) (C) \((1 / 2)\) (D) 1

The temperature of a substance increases by \(27^{\circ} \mathrm{C}\) What is the value of this increase of Kelvin scale? (A) \(300 \mathrm{~K}\) (B) \(2-46 \mathrm{~K}\) (C) \(7 \mathrm{~K}\) (D) \(27 \mathrm{~K}\)
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